pub fn new_birthday_probability(n: u32) -> f64 {
    // TODO: 这里写逻辑
    if n > 365 {
        return 1.0; // 当人数大于 365 时，一定有至少两人生日相同
    }

    let mut probability = 1.0;
    for i in 0..n {
        probability *= (365 - i) as f64 / 365.0;
    }

    let result = 1.0 - probability;
    (result * 10000.0).round() / 10000.0 // 保留四位小数
}
